An athlete on a trampoline leaps straight up into the air with an initial speed of 6.36 m/s. Find the maximum height reached by the athlete (in meters) relative to the trampoline. Use g=9.8 ms-2 and answer to 3 SF.

To find the maximum height reached by the athlete, we need to use the principles of motion and the laws of physics. We can start by determining the time it takes for the athlete to reach the maximum height.

We have the initial speed of the athlete, which is 6.36 m/s, and we know that at the maximum height, the velocity becomes zero (since the athlete will momentarily stop before falling back down). We can use the kinematic equation:

vf^2 = vi^2 + 2ad

Where:
- vf is the final velocity (0 m/s at the maximum height)
- vi is the initial velocity (6.36 m/s)
- a is the acceleration due to gravity (g = 9.8 m/s^2)
- d is the displacement

Rearranging the equation to solve for d:

d = (vf^2 - vi^2) / (2a)

Substituting the values, we have:

d = (0^2 - 6.36^2) / (2 * -9.8)

Simplifying:

d = (-40.4496) / (-19.6)
d = 2.0638 meters

Therefore, the maximum height reached by the athlete relative to the trampoline is approximately 2.06 meters.